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Figure 11 (a) Electricity supplied to homes has a frequency of □ A 0.02 Hz □ B 20 Hz □ C 50 Hz □ D 500 Hz (b) Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations - Edexcel - GCSE Physics - Question 5 - 2023 - Paper 2

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Figure-11--(a)-Electricity-supplied-to-homes-has-a-frequency-of--□-A-0.02-Hz-□-B-20-Hz-□-C-50-Hz-□-D-500-Hz--(b)-Explain-why-the-National-Grid-uses-high-voltages-with-small-currents-to-transfer-electricity-from-power-stations-Edexcel-GCSE Physics-Question 5-2023-Paper 2.png

Figure 11 (a) Electricity supplied to homes has a frequency of □ A 0.02 Hz □ B 20 Hz □ C 50 Hz □ D 500 Hz (b) Explain why the National Grid uses high voltages wit... show full transcript

Worked Solution & Example Answer:Figure 11 (a) Electricity supplied to homes has a frequency of □ A 0.02 Hz □ B 20 Hz □ C 50 Hz □ D 500 Hz (b) Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations - Edexcel - GCSE Physics - Question 5 - 2023 - Paper 2

Step 1

Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations.

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Answer

The National Grid uses high voltages for electricity transmission because it allows for the reduction of current flowing through the wires. Lower current reduces resistive losses (I²R losses) in the transmission lines, which can be significant when transferring large amounts of electricity over long distances. By operating at high voltages, the National Grid can efficiently transfer electricity from power stations to consumers while minimizing energy loss.

Step 2

Calculate the power in the primary coil.

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Answer

To calculate the power in the primary coil, we use the formula:

P=VimesIP = V imes I

Here, we substitute the voltage (assumed to be either given or relevant for your case) and the current accordingly based on the provided values. If, for instance, we consider a voltage of 80000 V and a current of 0.02 A, the calculation becomes:

P=80000imes0.02=1600 WP = 80000 imes 0.02 = 1600 \text{ W}

Thus, the power in the primary coil would be 1600 W.

Step 3

For the transformer in Figure 12, evaluate, in its simplest form, the ratio secondary voltage : primary voltage.

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Answer

For a transformer, the voltage ratio is based on the number of turns in the coils. If given the turns ratio as a fraction, if there are, for example, 1600 turns in the secondary coil and 80000 turns in the primary coil, the ratio can be calculated as:

Ratio (secondary voltage : primary voltage)=160080000=150\text{Ratio (secondary voltage : primary voltage)} = \frac{1600}{80000} = \frac{1}{50}

Thus, in its simplest form, the ratio of secondary voltage to primary voltage is 1:50.

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